An example for a calculation where imaginary numbers are used but don't occur in the question or the solution.

In a presentation I will have to give an account of Hilbert's concept of real and ideal mathematics. Hilbert wrote in his treatise "Über das Unendliche" (page 14, second paragraph. Here is an English version - look for the paragraph starting with "Let us remember that we are mathematicians") that this concept can be compared with (some of) the use(s) of imaginary numbers.

He thought probably of a calculation where the setting and the final solution has nothing to do with imaginary numbers but that there is an easy proof using imaginary numbers. I remember once seeing such an example but cannot find one, so:

Does anyone know about a good an easily explicable example of this phenomenon?

("Easily" means that enigneers and biologists can also understand it well.)

The canonical example seems to be Cardano's solution of the cubic equation, which requires non-real numbers in some cases even when all the roots are real. The mathematics is not as hard as you might think; and as an added benefit, there is a juicy tale to go with it – as the solution was really due to Scipione del Ferro and Tartaglia.

Here is a writeup, based on some notes I made a year and a half ago:

First, the general cubic equation $x^3+ax^2+bx+c=0$ can be transformed into the form $$ x^3-3px+2q=0 $$ by a simple substitution of $x-a/3$ for $x$.

We may as well assume $pq\ne0$, since otherwise the equation is trivial to solve.

So we substitute in $$x=u+v$$ and get the equation into the form $$ u^3+v^3+3(uv-p)(u+v)+q=0. $$ Now we add the extra equation $$ uv=p $$ so that $u^3+v^3+q=0$. Substituting $v=p/u$ in this equation, then multiplying by $u^3$, we arrive at $$ u^6+2qu^3+p^3=0, $$ which is a quadratic equation in $u^3$. Noticing that interchanging the two roots of this equation corresponds to interchanging $u$ and $v$, which does not change $x$, we pick one of the two solutions, and get: $$ u^3=-q+\sqrt{q^2-p^3}, $$ with the resulting solution $$ x=u+p/u. $$ The three different cube roots $u$ will of course yield the three solutions $x$ of the original equation.

Real coefficients

In the case when $u^3$ is not real, that is when $q^2<p^3$, we could write instead $$ u^3=-q+i\sqrt{p^3-q^2}, $$ and we note that in this case $\lvert u\rvert=\sqrt{p}$, so that in fact $x=u+\bar u=2\operatorname{Re} u$. In other words, all the roots are real.

In fact the two extrema of $x^3-3px+2q$ are at $x=\pm\sqrt{p}$, and the values of the polynomial at these two points are $2(q\mp p^{3/2})$. The product of these two values is $4(q^2-p^3)<0$, which is another way to see that there are indeed three real zeros.

Well, you can consider this sequence of integers: $$ u_0 = 1; u_1 = 1; u_{n+2} = u_{n+1} - u_n $$

This recurring definition is closely linked to the equation: $$ x^2 = x - 1 \Leftrightarrow x^2 - x + 1 = 0 $$ The solutions in $\mathbb{C}$ are $\frac{1 \pm i\sqrt3}{2}$ (complex numbers), and you can easily prove by recursion that: $$ u_n = \left(\frac{1 + i\sqrt3}{2}\right)^n + \left(\frac{1 - i\sqrt3}{2}\right)^n $$

which is an... integer ! So yes, you can have complex numbers that ease calculations of totally non-complex problems.